If $x \dagger y = 3x-8y$ and $x \star y = 4x^{2}+y^{2}$, find $(0 \star -1) \dagger -2$.
Answer: First, find $0 \star -1$ $ 0 \star -1 = 4(0^{2})+(-1)^{2}$ $ \hphantom{0 \star -1} = 1$ Now, find $1 \dagger -2$ $ 1 \dagger -2 = (3)(1)-(8)(-2)$ $ \hphantom{1 \dagger -2} = 19$.